In triangle $ABC$, angle $ACB$ is 50 degrees, and angle $CBA$ is 70 degrees.  Let $D$ be the foot of the perpendicular from $A$ to $BC$, $O$ the center of the circle circumscribed about triangle $ABC$, and $E$ the other end of the diameter which goes through $A$.  Find the angle $DAE$, in degrees.

[asy]

unitsize(1.5 cm);

pair A, B, C, D, E, O;

A = dir(90);

B = dir(90 + 100);

C = dir(90 - 140);

D = (A + reflect(B,C)*(A))/2;

E = -A;

O = (0,0);

draw(Circle(O,1));

draw(A--B--C--cycle);

draw(A--D);

draw(A--E,dashed);

label("$A$", A, N);

label("$B$", B, W);

label("$C$", C, SE);

label("$D$", D, SW);

label("$E$", E, S);

dot("$O$", O, dir(0));

[/asy]
Since triangle $ACD$ is right, $\angle CAD = 90^\circ - \angle ACD = 90^\circ - 50^\circ = 40^\circ$.

[asy]
unitsize(2 cm);

pair A, B, C, D, E, O;

A = dir(90);
B = dir(90 + 100);
C = dir(90 - 140);
D = (A + reflect(B,C)*(A))/2;
E = -A;
O = (0,0);

draw(Circle(O,1));
draw(A--B--C--cycle);
draw(A--D);
draw(A--E);
draw(O--C);

label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, SE);
label("$D$", D, SW);
label("$E$", E, S);
dot("$O$", O, NE);
[/asy]

Also, $\angle AOC = 2 \angle ABC = 2 \cdot 70^\circ = 140^\circ$.  Since triangle $ACO$ is isosceles with $AO = CO$, $\angle CAO = (180^\circ - \angle AOC)/2 = (180^\circ - 140^\circ)/2 = 20^\circ$.  Hence, $\angle DAE = \angle CAD - \angle CAO = 40^\circ - 20^\circ = \boxed{20^\circ}$.